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Differential Equations: Solution Differential Equation

JEE Maths question with a full step-by-step solution.

Question
The solution of the differential equation (x2sin⁡3y−y2cos⁡x) dx+(x3cos⁡ysin⁡2y−2ysin⁡x) dy=0(x^2\sin^3 y - y^2\cos x)\,dx + (x^3\cos y\sin^2 y - 2y\sin x)\,dy = 0 is:
Ax3sin⁡3y=3y2sin⁡x+Cx^3\sin^3 y = 3y^2\sin x + Ccorrect
Bx3sin⁡3y+3y2sin⁡x=Cx^3\sin^3 y + 3y^2\sin x = C
Cx2sin⁡3y+y3sin⁡x=Cx^2\sin^3 y + y^3\sin x = C
D2x2sin⁡y+y2sin⁡x=C2x^2\sin y + y^2\sin x = C
Solution
Group the terms so that each group is an exact differential:
(x2sin⁡3y dx+x3sin⁡2ycos⁡y dy)−(y2cos⁡x dx+2ysin⁡x dy)=0.\big(x^2\sin^3 y\,dx + x^3\sin^2 y\cos y\,dy\big) - \big(y^2\cos x\,dx + 2y\sin x\,dy\big) = 0.
 Now note
d(x33sin⁡3y)=x2sin⁡3y dx+x3sin⁡2ycos⁡y dy,d\left(\frac{x^3}{3}\sin^3 y\right) = x^2\sin^3 y\,dx + x^3\sin^2 y\cos y\,dy,
d(y2sin⁡x)=y2cos⁡x dx+2ysin⁡x dy.d\big(y^2\sin x\big) = y^2\cos x\,dx + 2y\sin x\,dy.
 So the equation is d(x33sin⁡3y)−d(y2sin⁡x)=0d\left(\dfrac{x^3}{3}\sin^3 y\right) - d\big(y^2\sin x\big) = 0, giving
x33sin⁡3y−y2sin⁡x=c⇒x3sin⁡3y=3y2sin⁡x+C.\frac{x^3}{3}\sin^3 y - y^2\sin x = c \Rightarrow x^3\sin^3 y = 3y^2\sin x + C.
 Correct answer: (1)
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